For a motor's output power to equal its input power, the motor's efficiency must be 100%. As no machine, particularly a rotating machine, can possibly achieve 100% efficiency, there is no condition under which its output power can ever match its input power.
The FLA of a motor is the amperage that the motor operates at. So to answer the question, 100%.
For AC system speed of generator will be decreased and then frequency decrease, if there is no trip the motor load will be automaticaly reduced as Load Power = torque x speed and speed of motor vary acc to frequency. Finally the motor will be run at new lower speed that load = power generation. The other effect is voltage in the system will be decreased and result of lower load consumption
The full load power expression in an induction motor is given by the formula: ( P_{out} = \eta \cdot P_{in} ), where ( P_{out} ) is the output power (mechanical power delivered by the motor), ( \eta ) is the efficiency of the motor, and ( P_{in} ) is the input power (the electrical power supplied to the motor). Additionally, the input power can be calculated as ( P_{in} = V \cdot I \cdot \sqrt{3} \cdot \cos(\phi) ) for three-phase motors, where ( V ) is the voltage, ( I ) is the current, and ( \cos(\phi) ) is the power factor. This expression accounts for real power consumption and efficiency at full load conditions.
The no-load current of a motor, such as a 90 kW motor operating at 440V and 60Hz, can vary based on its design and efficiency. Typically, the no-load current for such motors ranges from 10% to 30% of the full-load current. To estimate the no-load current, you can use the formula: No-load current ≈ Full load current × (no-load current percentage). The full-load current can be calculated using the formula: Full Load Current (A) = Power (W) / (Voltage (V) × √3 × Power Factor).
because of high resistance in the load.
Yes. A motor is considered to be a load of the power supply in use.
For a purely resistive load with a unity power factor, 9.41 kVA would equal 9.41 kW. However some equipment such as a motor will have a power factor less than 1. If the power factor is 0.8 then 9.41 kVA would equal 9.41 x 0.8 kW.
It is maximum at about 75% to 100% of the motor rated load. Efficiency is maximum at unity power factor , when R=X and when variable losses Is equal to constant losses at rated load.
The FLA of a motor is the amperage that the motor operates at. So to answer the question, 100%.
Ideally all three phase currents will be equal. There may exist some voltage imbalance from the power source, which will result in unequal currents.
torque load, generation load, power correction load
A:is equal to the source
For AC system speed of generator will be decreased and then frequency decrease, if there is no trip the motor load will be automaticaly reduced as Load Power = torque x speed and speed of motor vary acc to frequency. Finally the motor will be run at new lower speed that load = power generation. The other effect is voltage in the system will be decreased and result of lower load consumption
dun know
The full load power expression in an induction motor is given by the formula: ( P_{out} = \eta \cdot P_{in} ), where ( P_{out} ) is the output power (mechanical power delivered by the motor), ( \eta ) is the efficiency of the motor, and ( P_{in} ) is the input power (the electrical power supplied to the motor). Additionally, the input power can be calculated as ( P_{in} = V \cdot I \cdot \sqrt{3} \cdot \cos(\phi) ) for three-phase motors, where ( V ) is the voltage, ( I ) is the current, and ( \cos(\phi) ) is the power factor. This expression accounts for real power consumption and efficiency at full load conditions.
The difference in voltage is a minor detail and the power used depends largely on the mechanical load power on the motor.
When you do a load test on a 3-phase induction motor you are checking the power factor, viz slip, and efficiency of the motor. You can test various loads with this test.